Question

What volume of 0.6 m hcl is required to neutralize 100.0 ml of 0.5 m naoh?

Answer 1

Explanation:

The given data is as follows.

      $M_{1}$ = 0.6 M,          $V_{1}$ = ?

      $M_{2}$ = 0.5 M,         $V_{2}$ = 100.0 ml

Therefore, we calculate the volume of NaOH as follows.

   $(Molarity \times Volume)_{HCl}$ = $(Molarity \times Volume)_{NaOH}$

        $M_{1} \times V_{1} = M_{2} \times V_{2}$

                  $V_{1} = \frac{M_{2} \times V_{2}}{M_{1}}$

                                   = $\frac{0.5 M \times 100.0 ml}{0.6 M}$

                                   = 83.33 ml

Thus, we can conclude that volume of HCl is 83.33 ml.

Answer 2

In order to determine the amount needed, we have to know the chemical reaction that would happen between the substances. In this case the reaction would be:

NaOH + HCl = NaCl + H2O

We calculate as follows:

.1 L (0.5 mol NaOH / L ) ( 1 mol HCl / 1 mol NaOH) ( 1 L / 0.6 mol HCl ) = 0.08 L or 80 mL of 06 M HCl

Hope this answers the question. Have a nice day.